Allright within all your explanations the magenta vector is my orbital vector.

Concerning my coordinate you still don't seem to understand why I'm using it.

I don't calculate orbits or ellipses but directions of movements and the speed of a vehicle only. To do this I do need only one location of Earth as the planet that the vehicle will be launched from - and I need a scale to measure directions and speeds by. The vehicle is getting the initial direction and speed of Earth at launch. So I use the coordinate system as a representation of the vehicle - and as long as the vehicle isn't launched this coordinate system is representing Earth too. The coordinate system is moving and - as long as the vehicle isn't launched - orbiting sun. (I wonder that you didn't recognise this.)

It may be that I sometimes use a second coordinate system representing Earth instead of the already launched vehicle.

So far an explanation in principle.

The movement of my coordinate systems means that - different to your diagrams - the course of the vehicle (or in the case of the second coordinate system the orbit of Earth) a little bit is serving as a third coordinate TIME.

The orbital velocity changes in an elliptical orbit according to Kepler’s 2nd law. The velocity at perihelion is faster than the velocity at aphelion. Both the tangential and orbital velocities do this. And they change only due to the gravity of the Sun. Yet in your calculations you never let the tangential velocity change. How can you get a correct elliptical orbit if the tangential velocity does not change?

don't forget that I use a circular orbit as an approximation because Earth's orbit has such a low excentricity that it is not far from a circle - I already said that repeatedly.

The second point is that I do make use of Earth's velocity only at the time of launch of the vehicle. After that launch it would be easy to establish a circular orbit of the vehicle because of the steerability and controlability of the vehicle by its pulsed fusion engine and other technologies possible today.

Along a circular orbit velocity isn't changing.

Your remark is sounding a little bit like you forgot again an element of my concept.

And I'm not doing general considerations of orbits - I only do calculations of the movements and course of a vehicle launched from Earth going closer to the sun to throw a package of radioactive materials into the sun. Orbits that have a larger excentricity tahn Earth's orbit don't be of any interest because the package of radioactive materials is to be launched from Earth nad not from any other planet or body. So I leave aside all the mor excentrical ellipses and orbits - please do so too. I'm not calculating orbits of planets - use of Earth's orbit I made only to get the strating values for the vehicle to be launched.

And please keep in mind - I'm not working on a proposition or suggestion for reality but analysing something by a mathematical model which is a close APPROXIMATION for simplicity. (All this I already said several times during this thread and the other).

You must take all my explanation as a whole thing to understand why I'm doing what I'm doing and my doings themselves. The APPROXIMATION can be adjusted closer to reality after my calculations are ready.

Orbits that have a larger excentricity tahn Earth's orbit don't be of any interest

That is why your description of how you calculate the path is not adequate. Because your only explanation IN DETAIL has been the first second, and it was based on a circle.

Ekkehard Augustin wrote:

I'm not calculating orbits of planets

Yes you are! AFTER it is dropped, the package of waste will move exactly like a small planet. Actually more like a comet. But comets and planets are both obeying the same laws of Physics. The ONLY difference is that planets are in NEARLY circular orbits and comets are in VERY eccentric orbits. And comets that hit the sun have their very eccentric orbits interrupted by the surface of the Sun.

I have written a new Java program. Please try it and let me know how it works for you. It runs very fast some on computers and very slow on others, probably due to the version of Java installed.
http://home.austin.rr.com/campbelp/orbit.htmlIt even has a button to drop the package of waste, so you can maneuver the rocket, drop the package and maneuver the rocket again. If you point the rocket against the orbital direction, enter 26.0 in the DeltaV (km/s) field and click the Fire button, most of the orbital velocity will be cancelled and it will hit the Sun. You may want to slow down the simulation to allow time for rotating and firing and then speed it up again to see the result quickly
If you point the rocket directly toward the Sun and fire with DeltaV of 26.0, you will start moving closer the Sun but you will never get anywhere near it because your new velocity still has a component from the original orbital velocity which has not been cancelled. And after you miss the Sun you are in an Earth crossing orbit that reaches out past Jupiter.

At the default speed of “1”, each calculation represents about 2 minutes. Each click of the Slower button cuts that in half and each click of the Faster button doubles it. So 7 clicks slower will give you about 1 second per calculation. At the default speed running on my fastest (best Java) computer it completes one orbit in about 35 seconds. At the one calculation per second speed it is 128 times slower. But you really don’t need it that slow to get results accurate enough for our discussion.

The MOST EFFICIENT way to hit the Sun is to cancel most of the Earth’s velocity with thrust directed against the Earth’s orbital velocity.

The LEAST EFFICIENT way to hit the Sun is to thrust directly at the Sun, because the new velocity will still have a component from the original orbital velocity at right angles to the intended direction of motion which causes you to miss the Sun, unless an EXTREMELY large velocity (many times larger than solar escape velocity) is used.

And hitting the Sun using even the most efficient method (canceling the Earth’s orbital motion) still takes twice as much energy as hitting Pluto would require.

The Java program uses the same math as http://home.austin.rr.com/campbelp/orbit.xlsOrbit.xls starts computing at an arbitrary position 10 units up the Y axis from the origin, where the Sun is assumed to be. If the initial speed is 0.31692 to the right then the result is a nearly circular orbit. Multiply the distance by 15,000,000 and the speed by 94.03 to scale it to distance and speed in the real solar system. With these starting conditions it simulates one full orbit in about 200 steps.

If you increase the initial distance by a factor of 4 and cut the initial speed in half, you get another nearly circular orbit calculated in smaller steps. You could consider it as the same orbit by multiplying the distance by a quarter as much (3,750,000) and the speed by half as much (47.015) to scale it to the real solar system.

If you change the start distance and speed as described above, the simulation will complete one orbit in about 1600 steps, or 8 times as many as before. You can clearly see Kepler’s 3rd law ( P^2 = A^3 ) in these data, proof that the simulation is correct. Try it, you will see. (Don’t forget to copy the last line 1200 more times and modify the graph to show all 1600 results)

If you reduce the initial velocity to 20 km/s, you get an elliptical orbit with perihelion at 44,000,000 km and speed of 68 km/s. You can clearly see Kepler’s 1st and 2nd laws (elliptical orbits and equal area in equal time) in these data. Another proof that the simulation is correct. Try it, you will see.

If you increase the initial speed by SQRT(2), you get a parabola that never returns. This is the SQRT(2) times circular orbit speed equals escape velocity rule. Yet another proof that the simulation is correct. Try it, you will see.

If you increase the initial speed even more you get increasingly flat hyperbolas escaping at higher velocity. Yet another proof that the simulation is correct. Try it, you will see.

If you calculate the orbital energy at various points for an elliptical orbit without any extra DeltaVs, it remains constant. The orbital energy for the parabolic escape orbit is 0, as it should be. Hyperbolic escape orbits have positive energy, as they should. More proof that the simulation is correct. Try it, you will see.

If you apply a DeltaV that is not tangent to the orbit you may be confused by the result. It behaves very much like a gyroscope. This is the correct result because orbits and gyroscopes are both rotating dynamical systems. Such systems behave in ways that are guaranteed to confuse. Such behavior is EXTREMELY difficult to justify with only vector calculations such as we are doing, but it is possible, as orbit.xls shows.

I have written versions of this simulator that take the gravity of many bodies into account. Each object has gravity and mass so that all the objects accelerate each other. With this expanded simulator I have correctly simulated the Moon orbiting the Earth while the Earth orbits the Sun. I have also simulated space craft gravity assist maneuvers.

All these proofs demonstrate that this simulation is correct within the limitations of any approximation of curved motion by small straight segments.

you wrote "AFTER it is dropped, the package of waste will move exactly like a small planet. Actually more like a comet."

Exactly this is incorrect in the case of the vehicle I am using - what you wrote is valid in the case of a vehicle that is launched the conventional way until it has escape velocity and then never again will be accelerated or propelled artificially.

What you wrote is made use of concerning all vehicles using a chemical engine because such an engine requires such a large amount of propellant for accelration and control of the course that it is impossible to make it work permanently. For this reason conventional interplanetary vehicles are using orbital mecahnics to be able to reach Saturn, Titan, Jupiter, Mars, Eros or comets.

For this reason I don't use a vehicle with a chemical drive - I allready said that the vehicle I'm using in my mathematical model has a pulsed fusion engine. A pulsed fusion engine can work permanently without requiring that huge amount of propellant that a chemical engine would require. That's a SIGNIFICANT difference compared to all vehicles NASA has sent to other bodies in the past (except Deep Space One which had a permanently working ion engine) The significant difference between chemical engines, ion engines and the pulsed fusion engine is their efficiency abd the impulse got by them.

And at the first second the vehicle in my model is accelerated by an additional concept which too is much more efficient than a chemical engine and requires much less propellant than a chemical engine - it requires NO propellant - : the space elevator which is giving the vehicle an initial velocity of more than 7 km/s to the direction desired.

The 7 km/s got by the space elevator is very much more than the acceleration by sun's gravity which is less than a meter per second - the 1 g that can be got by the pulsed fusion engine is much more than the acceleration by sun's gravity too.

For this reason I first can look what will happen if the vehicle is accelerated by the elevator during in the first second only and then is mving without artificial acceleration and propulsion for a while. The fact that I do accelerate the vehicle directly towards the sun in the first second but not towards the direction of Earth's tangential vector means that I do not change the vehicle's tangential vector - the vehicle still has the same tangential vector as Earth. But the acceleration by the space elevator is giving the vehicle a much larger vertical vector (towards sun) than the vertical vector of Earth. This is valid in the first second. Because of the difference in the vertical vector the resulting vector of the vehicle is different to Earth's resulting vector (your magenta vector) - and Earth's resulting vector is Earth's orbital vector (your magenta vector). The magenta vector of the vehicle is different then to Earth's magenta vector both in direction and length. Because the vehicle has no touch and no connection to Earth then direction and velocity of Earth don't matter for the vehicle after the launch.

The acceleration by the space elevator in the first second is an artificial external acceleration which means that the vehicle is independent of the forces involved in orbital mechanics in that first second. If after this first second no artificial external acceleration and propulsion is working then the vehicle is obeying orbital mechanics until the pulsed fusion engine is fired. And this I have been calculating in my Step Two in the other discussion - you calculated it your way and got a result not that extreme different to mine. And I recognized an error in my calculation which I removed - I will have to continue my recalculation to get the new result as basis for the third step and as basis for comparison to your own result for Step Two.

Each tima I use the engine of my vehicle the vehicle is independent of orbital mechanics - orbital mechanics have an impact on the vehicle as they have on each body but the engines of the vehicle are that crafty that they are easyly overriding orbital mechanics. An acceleration of 1 g is much more than the acceleration by sun's gravity at the distance of Earth.

What you say is valid in the case of chemical engines but I don't use them - so don't argue based on them. It is invalid to argue on the basis of chemical engines against the working of non-chemical engines or pulsed fusion engines. The two have different physics and different capabilities.

We are calculating 2 paths. Path 1 is the powered vehicle carrying the package. Path 2 is the path of the package after it is released. Path 2 WILL obey orbital mechanics because the package has no engine. I don’t care how you get to the point of releasing the package. Just tell me the location where you release it and velocity it has when you release it, and I will tell you if the package will hit the Sun or not.

At the moment it is released, the package is at point A with an instantaneous velocity it got from the space craft. This instantaneous velocity is shown by vector Black 1. At point A the gravitational acceleration of the Sun is Green 1, which is calculated from 1/R^2, where R is the distance to the Sun.
To determine where the package will go in the first second, complete the parallelogram made by vectors Black 1 and Green 1 to get vector Magenta 1. Then ASSUME the package moves from point A to point B along vector Magenta 1.

The package arrives at point B with instantaneous velocity Magenta 1. Copy Magenta 1 to Black 2. Black 2 has the same direction and length as Magenta 1, but it starts at point B. Now at the end of the first second the package is at point B with instantaneous velocity Black 2 and feeling a different gravitational acceleration shown by Green 2. Green 2 is calculated from 1/R^2, but this R is different than at point A.

Repeat the same calculation (complete the parallelogram) at point B using Black 2 and Green 2 to get Magenta 2. ASSUME the package moves from point B to point C along Magenta 2.

The package arrives at point C with instantaneous velocity Magenta 2. Magenta 2 = Black 3. Calculate Green 3 at point C and complete the parallelogram with Black 3 and Green 3 to get Magenta 3.

ASSUME the package moves to point D along Magenta 3. And so on.

Now we are assuming that the package moves along the Magenta vectors, but it doesn’t really. It moves along a continuously curving path with a velocity that starts equal to the black vector and smoothly changes to the magenta vector as it moves. But this approximation is good enough if the vectors are all short enough. And the result would be exactly correct if the vectors were infinitely short.

Do we agree that this is the correct way to calculate the path of an object moving freely in space and being controlled only by the gravity of the Sun?

And to calculate the path of the powered vehicle, calculate the direction and size of the thrust at each point and ADD it to the Green gravity vectors with a parallelogram. Then use the modified Green vector in the calculation above. The thrust does NOT remove gravity. It merely is added to it and overwhelms it.

So when the engine is operating, the Green vector is 1/R^2 + thrust. Then calculate as above.

I cannot answer to your whole post this moment and will do that later - tomorrow perhaps- but I'm not calculating two paths that way.

I first calculate the path of the vehicle until a special situation at a special point occurs. Then the package will be released - in that special situation at that special point. Then and only then I will calculate the path of the package. But before the package will go the same path as the vehicle and both parts of the path of the package are forming the whole path of the vehicle. And the question that caused this discussion requires the calculation of this whole path. So I consequently will calculate the whole path and not the path of the package after its release only.

The vehicle hasn't reached that special point yet, the package isn't released yet and so none of these both are without an engine yet.

If you are calculating the package after it has been released then you are calculating something currently not under consideration by me .

If you have not even considered that path of the package after it is released, how do you know it would hit the Sun?

Ekkehard Augustin wrote:

at that special point

What special point? What distance from the Sun? What velocity and in what direction will the package be travelling? That is all I need to know to determine if the package will hit the Sun. You don’t need to calculate how to get to this special point, just tell me where it is!

By the way, have you tried the Java program?
http://home.austin.r.com/campbelp/orbit.htmlIt works fine on some computers and VERY slow on others. I haven't figured out why yet. A good computer will show 1 complete orbit in under a minute at the default speed.

I prefer to let the vehicle move second by second, to look at distance, gravitational acceleration and direction after each second and then to decide waht to do next. Gravitational acceleration is increasing by square each second.

The calculation of step two provides a comparison between distance gone if there were no gravitational acceleration and distance gone in presence of gravitational acceleration.

I want to act if I were the captain of the vehicle having no shape of the path or course in mind. All decisions I want to base upon a look at the actual situation - including the decison wether or not to release the package. Currently I have to correct my calculations.

I will let the vehicle move its way and do calculations for a released package only when the point of release is reached. I have several reasons for this - one but only one of them is to keep my overview about the whole process.

I've read in total your post now you wrote yesterday but still cannot answer to it in whole. But one thing I want to explain. In the second the vehicle is released by the space elevator the gravitational acceleration and the velocity of the released vehicle are point to exactly the same direction at the same time. This way the vehicle is getting a vertical vector with a velocity which is the sum of the more than 7 km/s got by the elevator and the gravitational acceleration in that second which is 0.000000592 km/s according to "Schülkes Tafeln". By this I get the green vector at the end of the first second. The elevator is giving thrust and impulse to the vehicle.

Beginning in the scond second I have to reduce the velocity got by the elevator because this fraction of total vertical velocity no longer is pointing to exactly the same direction as the gravitational acceleration. The new total vertical velocity is slightly less than in the first second - but only towards the center of sun. No new thrsut is given in this second and so no correction for pointing to the center of sun - which is the source of gravitation - is done.

The special situation has to do with the permanent change of gravitational acceleration and artificial accelöeration I'm not giving during the second step of my calculations.

When the package is released I have to calculate two paths in parallel - the path of the vehicle and the path of the package. I have to keep in mind that the vehicle shouldn't fall into the sun but return to Earth.

Please keep in mind that the vehicle isn't on a scientific mission but on a commercial one - decsions are not to be optimized under physical or technical aspects only and may be suboptimal seen from Physics and Technics.

I'm surprised. The actual situation simply and clearly is described by the distance got after the last calculated (or flown) second, the gravitational acceleration at that distance, the direction the resulting (magenta) vector is pointing to and the vertical vector valid there.

And - it's an actual situation I'm faced to: No need to look for it.

The decisions to be based on this during the second step and at the beginning of the third step don't have anything to do with the release of the package in the special situation.

The special situation I have in mind already is well defined - but the vehicle still has to go a little bit before this situation is valid and I want to use this time to do some useful decisions about artificial acceleration, desried direction etc.